THE SUMS OF SQUARE TECHNIQUE
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Võ Quc Bá Cn Phm Th Hng
Copyright © Vo Quoc Ba Can 1
THE SUMS OF SQUARE TECHNIQUE
I. Theorem.
Consider the following inequality
422332
( )0
cyc cyc cyc cyc cyc
m a n a b p a b g ab m n p g a bc
+++−+++≥
∑∑∑∑∑
With
,,
abc
be real numbers.
Then this inequality holds when
22
0
3()
m
mmn p pgg
>
+≥++
.
Proof.
We rewrite the inequality as
422 222 3 2
32
()
0
cyc cyc cyc cyc cyc cyc
cyc cyc
m a ab m n ab abc p ab abc
g ab a bc
−++ −+ −
+ −≥
∑∑ ∑ ∑ ∑∑
∑∑
Note that
4 2 2 2 22
3 2 3 2 22
22 22 22
3 2 3 2 22
22
1
()
2
()
11
()( )()()( 2)
33
()
()
cyc cyc cyc
cyc cyc cyc cyc cyc
cyc cyc cyc
cyc cyc cyc cyc cyc
cy
a ab ab
ab abc bc abc bca b
bc a b ab bc ca a b a b ab ac bc
ab a bc ca ab c ca a b
caab
−=−
−=−= −
=− − + ++ − = − +−
−=−= −
=−
∑∑∑
∑∑∑∑∑
∑ ∑∑
∑∑∑∑∑
22 22
11
( ) ( ) ( )( 2)
33
c cyc cyc
ab bc ca a b a b ab bc ca
− ++ − =− − +−
∑ ∑∑
Then the inequality is equivalent to
222 22
222
1
( ) ( )[()(2)(2)]
23
()0
cyc cyc
cyc cyc
m
a b a b p g ab p g bc p g ca
m n a b a bc
− + − − − + ++
++ −≥
∑∑
∑∑
Moreover
2222
22
1
[()(2)(2)]
6()
cyc cyc cyc
a b a bc p g ab p g bc p g ca
p pgg
− = − − + ++
++
∑∑∑
The inequality becomes
Võ Quc Bá Cn Phm Th Hng
Copyright © Vo Quoc Ba Can 2
222 22
2
22
222
1
( ) ( )[()(2)(2)]
23
[( ) (2 ) (2)]0
6()
1
[3( )()(2)(2)]
18
cyc cyc
cyc
cyc
m
a b a b p g ab p g bc p g ca
mn
p g ab p g bc p g ca
p pgg
ma b p gab p gbc p gca
m
− + − − − + ++
+
+ −−+++≥
++
⇔ − +− − + ++
∑∑
∑
∑
22
2
22
3()
[()(2)(2)]0
18()
cyc
mmn p pgg
p g ab p g bc p g ca
mp pgg
+−−−
+ −−+++≥
++
∑
From now, we can easily check that if
22
0
3()
m
mmn p pgg
>
+≥++
then the inequality is true.
Our theorem is proved. J
II. Application.
Example 1
. (Vasile Cirtoaje) Prove that
2222 333
( ) 3( ).
a b c ab bc ca
++ ≥ ++
Solution.
The inequality is equivalent to
4 223
20
cyc cyc cyc
a ab ab
+ −≥
∑∑∑
From this, we get
1,2, 3,0
mnpg
== =−=
, we have
22 22
10
3 ( ) 3 1 (1 2) ( 3) ( 3) 0 0 0
m
mmn p pgg
=>
+ − − − =⋅⋅ + −− −−⋅− =
Then using our theorem, the inequality is proved. J
Example 2
. (Võ Quc Bá Cn) Prove that
( )
444 333
3 1 ( ) 3( ).
a b c abca b c ab bc ca
+++ − ++≥ ++
Solution.
We have
1,0, 3,0
mnpg
===−=
and
( ) ( )
2
222
10
3( ) 31(10) 3 3000
m
mmn p pgg
=>
+ − − − = ⋅⋅ + −− −− ⋅ − =
Then the inequality is proved. J
Example 3
. (Phm Vn Thun) Prove that
444 333
7( ) 10( ) 0.
abc abbcca
+++ ++≥
Solution.
We will prove the stronger result, that is
4
43
17
7 10
27
cyc cyc cyc
aaba
+≥
∑∑∑
Võ Quc Bá Cn Phm Th Hng
Copyright © Vo Quoc Ba Can 3
4223 32
86 51 101 34 102 0
86
51
101
34
cyc cyc cyc cyc cyc
a a b a b ab a bc
m
n
p
g
⇔−+−−≥
=
=−
⇒
=
=−
∑∑ ∑∑∑
Moreover
22 22
860
3 ( ) 3 86 (86 51) 101 101 ( 34) ( 34) 1107 0
m
mmn p pgg
=>
+−−−=⋅⋅−− − ⋅−−− = >
Then the inequality is proved. J
Example 4
. (Vình Quý) Let
, , 0, 1.
a b c abc
>=
Prove that
2 22
111
3.
111aa bb cc
++≤
−+ −+ −+
Solution.
On Mathlinks inequality forum, I posted the following proof:
Lemma. If
, , 0, 1,
a b c abc
>=
then
222
111
1.
111aa bb cc
++≥
++ ++ ++
Proof. From the given condition
,,0,1
a b c abc
>=
, there exist
,,0
xyz
>
such that
2
2
2
.
yz
a
x
zx
b
y
xy
c
z
=
=
=
And
then, the inequality becomes
4
4 2 22
1
cyc
x
x xyz yz
≥
++
∑
By the Cauchy Schwarz Inequality, we get
2 22
2 22
4
4 2 22
4 2 22 4 22 2 4 22
1
()2
cyc cyc cyc
cyc
cyc cyc cyc cyc cyc cyc
x xx
x
x xyz yz
x xyz yz x yz xyz x yz
≥ = ≥=
++
+++++
∑ ∑∑
∑
∑ ∑∑ ∑ ∑∑
Our lemma is proved.
Now, using our lemma with note that
222
222
111
,,0
,
111
1
abc
abc
>
⋅⋅=
we get
Võ Quc Bá Cn Phm Th Hng
Copyright © Vo Quoc Ba Can 4
422
42 42 42
22
22 22
1 2( 1)
124
111
( 1) ( 1) 1 1
44
( 1)( 1) 1 1
cyc cyc cyc
cyc cyc cyc
xxx
xx xx xx
xx xx
xx xx xx xx
++
≥⇔≤⇔≤
++ ++ ++
++ + −+
⇔ ≤⇔+≤
++ −+ −+ ++
∑∑∑
∑ ∑∑
Using our lemma again, we can get the result. J
Now, I will present another proof of mine based on this theorem
Since
, , 0, 1,
a b c abc
>=
there exists
,,0
xyz
>
such that ,,
yzx
abc
xyz
===
then our inequality
becomes
22 22
22 22 22 22
3 3 ( 2)
39433
cyc cyc cyc cyc
x x x xy
x xyy x xyy x xyy x xyy
−
≤⇔≤⇔−≥⇔≥
−+ −+ −+ −+
∑∑∑∑
By the Cauchy Schwarz Inequality, we get
2
2
2222
22
( 2)
( 2)( ) ( 2)
cyc cyc cyc
xy
xyxxyy xy
x xyy
−
− −+≥−
−+
∑∑∑
It suffices to show that
2
2 222
4 22 3 32
( 2) 3 ( 2)( )
10 39 25 16 8 0
cyc cyc
cyc cyc cyc cyc cyc
x y x yx xyy
x x y x y xy x yz
− ≥ − −+
⇔+ −−−≥
∑∑
∑∑ ∑ ∑∑
From this, we get
10, 39, 25, 16
mnpg
= = =−−
and
22 22
100
3 ( ) 3 10 (10 39) ( 25) ( 25) ( 16) ( 16) 189 0
m
mmn p pgg
=>
+ − − − = ⋅ ⋅ + −− −− ⋅− −− = >
Then using our theorem, the inequality is proved. J
III. Some problems for own study.
Problem 1
. (Vasile Cirtoaje) Prove that
444333 333
2( ).
a b c ab bc ca ab bc ca
+++++≥ ++
Problem 2
. (Phm Vn Thun, Võ Quc Bá Cn) Prove that
3334
8
( ) ( ) ( ) ( ).
27
aab bbc cca abc
+ + + + + ≥ ++
Problem 3
. (Phm Kim Hùng) Prove that
444 2 333
1
( ) 2( ).
3
a b c ab bc ca a b b c c a
+++ ++ ≥ + +
Võ Quc Bá Cn
Student
Can Tho University of Medicine and Pharmacy, Can Tho, Vietnam
E-mail:
Võ Quc Bá Cn Phm Th Hng
Copyright © Vo Quoc Ba Can 5
